1
$\begingroup$

Given the expression

$T = \max(|\beta_{\max}-1|,|\beta_{\min}-1|)$

Can we relate $T$ to variance $\sigma^2$ of values $\beta_i^2$, assuming we know $\sigma^2$, $\beta_{\max}$ and $\beta_{\min}$.

PS: What if values $\beta_i^2$ are distributed with mean $\mu=1$.

$\endgroup$
0

2 Answers 2

1
$\begingroup$

Let $b_i:=\beta_i$. Then $1-T\le b_i\le1+T$ for all $i$, and these inequalities provide the best bounds on the $b_i$'s given the information we have. So, $$A\le b_i^2\le B \tag{1}$$
for all $i$, where $$A:=\max[0,1-T]^2,\quad B:=(1+T)^2,$$ and inequalities (1) provide the best bounds on the $b_i^2$'s given the information we have.

So, for the variance $\sigma^2$ of the $b_i^2$'s we have $$0\le\sigma^2\le(B-\mu)(\mu-A), \tag{2}$$ where $\mu$ is the mean of the $b_i^2$'s. Moreover, inequalities (2) provide the best bounds on $\sigma^2$ given the information we have.

The latter two statements follow from

Lemma: Let $X$ be any random variable such that $EX=\mu$ and $A\le X\le B$ for some real $A$ and $B$. Then for the variance $\sigma^2$ of $X$ we have $$0\le\sigma^2\le(B-\mu)(\mu-A). \tag{3}$$ Moreover, inequalities (3) provide the best bounds on $\sigma^2$ in the conditions of this lemma.

Proof. Let $Y:=X-\mu$. Then $EY=0$ and $-a\le Y\le b$, where $b:=B-\mu$ and $a:=\mu-A$. So, $$0\le E(b-Y)(Y+a)=ba-EY^2=ba-\sigma^2.$$ So, $$\sigma^2\le ba=(B-\mu)(\mu-A),$$ so that (3) holds. Moreover, $\sigma^2=ba=(B-\mu)(\mu-A)$ if $Y$ takes values $-a,b$ with probabilities $\frac b{a+b},\frac a{a+b}$, respectively, that is, if $X$ takes values $A,B$ with probabilities $\frac b{a+b},\frac a{a+b}$, respectively.

$\endgroup$
2
  • $\begingroup$ @losif Thanks! Result in (2) is nice! But solving it for T, so as to express T in terms of $\sigma^2$ looks tricky because of max() involved. Any tips on that? $\endgroup$
    – Astro
    Jun 17, 2020 at 14:15
  • 1
    $\begingroup$ @Astro : If you want to solve the equation $(B-\mu)(\mu-A)=C$ for $T$ (for a given $C$), consider the two cases, $T\le1$ and $T>1$, separately. In the second case, the equation $(B-\mu)(\mu-A)=C$ is quadratic, and in the first case it is of degree $4$. This is what you have, anyway. $\endgroup$ Jun 17, 2020 at 15:37
0
$\begingroup$

If you know $\beta_{\max}$ and $\beta_{\min}$ then I would have thought you know $T$.

If you do not know $\beta_{\max}$ and $\beta_{\min}$ then you can say $T \ge \sigma$, with equality only when you have a distribution with half the probability at $1-\sigma$ and half the probability at $1+\sigma$ (and a mean of $1$)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.